This is a problem occurs to me when I was trying to find the limit of a function with two variables(if only it exists). Please help.
When I have to show that the limit does not exist for some function. Then by showing along two paths have two different limits I can prove it since the functions with two variables have infinite number of directions of approach to the given point.
So selecting the paths, should I only take the linear ones or can I take non linear paths also?
As an example,consider the limit of $\frac{xy^3}{x^2+y^6}$ as $(x,y)$ goes to $(0,0)$
Then taking along $x$ axis I have the limit $0$.
Can I take $y=x^{1/3}$ which gives me the limit $1/2$ ?
When I check the answer on some website it seems they haven't consider curved functions to find different limits.
Please explain. Where am I wrong?
Best Answer
You can take any path you want, as long as it actually helps you solve the problem. For example $$ \lim\limits_{(x,y)\to(0, 0)} \frac{xy^3}{x^2+y^6} $$ For the $y=x$ path, we have $$ \lim\limits_{(x,y)\to(0, 0)} \frac{xx^3}{x^2+x^6} = \lim\limits_{(x,y)\to(0, 0)} \frac{x^4}{x^2+x^6} = \lim\limits_{(x,y)\to(0, 0)} \frac{x^2}{1+x^4} = 0$$ For the $y=\sqrt[3]{x}$ path, we have $$ \lim\limits_{(x,y)\to(0, 0)} \frac{x\left(\sqrt[3]{x}\right)^3}{x^2+\left(\sqrt[3]{x}\right)^6} = \lim\limits_{(x,y)\to(0, 0)} \frac{x^2}{x^2+x^2} = \lim\limits_{(x,y)\to(0, 0)} \frac{1}{1+1} = \frac12$$ Therefore $$ \lim\limits_{(x,y)\to(0, 0)} \frac{xy^3}{x^2+y^6} \Rightarrow \mbox{does not exist}$$